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How do I show something is pathwise connected?

  1. Nov 1, 2011 #1
    I need to prove that X={(x,y):a<=x<=b, c<=y<=d}

    I was thinking of using proof by contradiction.

    Assume that X is not pathwise connected, then for a,b in X there is no continuous function that connects the two.

    I can show that then the set is disconnected but not sure where to go after that.

    Am I going about it the correct way?

    Thanks!
     
  2. jcsd
  3. Nov 1, 2011 #2

    micromass

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    Why can't you just give the path that connects two points??

    Given (x,y) and (u,v). What path can you take that connects (x,y) and (u,v)????
     
  4. Nov 1, 2011 #3
    if I take a straight line path from the points, it would be the distance between the two, so sqrt((x-u)^2 + (y-v)^2)
     
  5. Nov 1, 2011 #4

    micromass

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    I don't care about the distance. I want a function [itex][0,1]\rightarrow X[/itex]. A linear path is ok. How do you write that in function language?
     
  6. Nov 1, 2011 #5
    i am not sure what you mean by function language but I would want f(0) = (x,y) and f(1) = (u, v)
     
  7. Nov 1, 2011 #6

    micromass

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    Xhat is the equation of the line connecting your two points??
     
  8. Nov 1, 2011 #7
    i want to say something like

    given f(0) = (x, y) and p in [0,1] then f(p) = ((y-v)/(x-u)) * p + (x, y)

    but thats not correct bc i am switching between R and R^2

    do i need to make a continuous function for each component?
     
  9. Nov 1, 2011 #8

    micromass

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    Yes, you need to go linearly drom x to u and from y to v.
     
  10. Nov 2, 2011 #9
    do I do it peicewise? like f(p) = (x,y) for p=0, (u, v) for p=1 and then this other functions that I cannot figure out how to get.
     
  11. Nov 2, 2011 #10

    HallsofIvy

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    You need to prove what about this set? That it is connected? Or that it is pathwise connected? A set can be "connected" without being "pathwise connected".

     
  12. Nov 2, 2011 #11
    Pathwise connected
     
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